3.461 \(\int \frac{x^5 \sqrt{c+d x^3}}{(a+b x^3)^2} \, dx\)

Optimal. Leaf size=136 \[ \frac{\sqrt{c+d x^3} (2 b c-3 a d)}{3 b^2 (b c-a d)}-\frac{(2 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{5/2} \sqrt{b c-a d}}+\frac{a \left (c+d x^3\right )^{3/2}}{3 b \left (a+b x^3\right ) (b c-a d)} \]

[Out]

((2*b*c - 3*a*d)*Sqrt[c + d*x^3])/(3*b^2*(b*c - a*d)) + (a*(c + d*x^3)^(3/2))/(3*b*(b*c - a*d)*(a + b*x^3)) -
((2*b*c - 3*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/(3*b^(5/2)*Sqrt[b*c - a*d])

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Rubi [A]  time = 0.1095, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {446, 78, 50, 63, 208} \[ \frac{\sqrt{c+d x^3} (2 b c-3 a d)}{3 b^2 (b c-a d)}-\frac{(2 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{5/2} \sqrt{b c-a d}}+\frac{a \left (c+d x^3\right )^{3/2}}{3 b \left (a+b x^3\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*Sqrt[c + d*x^3])/(a + b*x^3)^2,x]

[Out]

((2*b*c - 3*a*d)*Sqrt[c + d*x^3])/(3*b^2*(b*c - a*d)) + (a*(c + d*x^3)^(3/2))/(3*b*(b*c - a*d)*(a + b*x^3)) -
((2*b*c - 3*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/(3*b^(5/2)*Sqrt[b*c - a*d])

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{x^5 \sqrt{c+d x^3}}{\left (a+b x^3\right )^2} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x \sqrt{c+d x}}{(a+b x)^2} \, dx,x,x^3\right )\\ &=\frac{a \left (c+d x^3\right )^{3/2}}{3 b (b c-a d) \left (a+b x^3\right )}+\frac{(2 b c-3 a d) \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{a+b x} \, dx,x,x^3\right )}{6 b (b c-a d)}\\ &=\frac{(2 b c-3 a d) \sqrt{c+d x^3}}{3 b^2 (b c-a d)}+\frac{a \left (c+d x^3\right )^{3/2}}{3 b (b c-a d) \left (a+b x^3\right )}+\frac{(2 b c-3 a d) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,x^3\right )}{6 b^2}\\ &=\frac{(2 b c-3 a d) \sqrt{c+d x^3}}{3 b^2 (b c-a d)}+\frac{a \left (c+d x^3\right )^{3/2}}{3 b (b c-a d) \left (a+b x^3\right )}+\frac{(2 b c-3 a d) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x^3}\right )}{3 b^2 d}\\ &=\frac{(2 b c-3 a d) \sqrt{c+d x^3}}{3 b^2 (b c-a d)}+\frac{a \left (c+d x^3\right )^{3/2}}{3 b (b c-a d) \left (a+b x^3\right )}-\frac{(2 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{5/2} \sqrt{b c-a d}}\\ \end{align*}

Mathematica [A]  time = 0.13852, size = 117, normalized size = 0.86 \[ \frac{\frac{(2 b c-3 a d) \left (\sqrt{b} \sqrt{c+d x^3}-\sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )\right )}{b^{3/2}}+\frac{a \left (c+d x^3\right )^{3/2}}{a+b x^3}}{3 b (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*Sqrt[c + d*x^3])/(a + b*x^3)^2,x]

[Out]

((a*(c + d*x^3)^(3/2))/(a + b*x^3) + ((2*b*c - 3*a*d)*(Sqrt[b]*Sqrt[c + d*x^3] - Sqrt[b*c - a*d]*ArcTanh[(Sqrt
[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]]))/b^(3/2))/(3*b*(b*c - a*d))

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Maple [C]  time = 0.01, size = 897, normalized size = 6.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(d*x^3+c)^(1/2)/(b*x^3+a)^2,x)

[Out]

1/b*(2/3*(d*x^3+c)^(1/2)/b+1/3*I/b/d^2*2^(1/2)*sum((-d^2*c)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)^(1/3)
+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(d*(x-1/d*(-d^2*c)^(1/3))/(-3*(-d^2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^(1/3))
)^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I
*(-d^2*c)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-d^2*c)^(2/3)+2*_alpha^2*d^2-(-d^2*c)^(1/3)*_alpha*d-(-d^2*c)^(2/3
))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))
^(1/2),1/2*b/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_alpha^2*d-I*(-d^2*c)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-d^2*c)
^(2/3)*_alpha-3*c*d)/(a*d-b*c),(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/
3)))^(1/2)),_alpha=RootOf(_Z^3*b+a)))-a/b*(-1/3*(d*x^3+c)^(1/2)/b/(b*x^3+a)-1/6*I/d/b*2^(1/2)*sum(1/(a*d-b*c)*
(-d^2*c)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(d*(x-1/d*(
-d^2*c)^(1/3))/(-3*(-d^2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d^2*c)^(1/3
)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-d^2*c)
^(2/3)+2*_alpha^2*d^2-(-d^2*c)^(1/3)*_alpha*d-(-d^2*c)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3
)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),1/2*b/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_alpha^2
*d-I*(-d^2*c)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*d)/(a*d-b*c),(I*3^(1/2)/d*(-d^2*c
)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*b+a)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(d*x^3+c)^(1/2)/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.84243, size = 707, normalized size = 5.2 \begin{align*} \left [-\frac{{\left ({\left (2 \, b^{2} c - 3 \, a b d\right )} x^{3} + 2 \, a b c - 3 \, a^{2} d\right )} \sqrt{b^{2} c - a b d} \log \left (\frac{b d x^{3} + 2 \, b c - a d + 2 \, \sqrt{d x^{3} + c} \sqrt{b^{2} c - a b d}}{b x^{3} + a}\right ) - 2 \,{\left (3 \, a b^{2} c - 3 \, a^{2} b d + 2 \,{\left (b^{3} c - a b^{2} d\right )} x^{3}\right )} \sqrt{d x^{3} + c}}{6 \,{\left (a b^{4} c - a^{2} b^{3} d +{\left (b^{5} c - a b^{4} d\right )} x^{3}\right )}}, \frac{{\left ({\left (2 \, b^{2} c - 3 \, a b d\right )} x^{3} + 2 \, a b c - 3 \, a^{2} d\right )} \sqrt{-b^{2} c + a b d} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-b^{2} c + a b d}}{b d x^{3} + b c}\right ) +{\left (3 \, a b^{2} c - 3 \, a^{2} b d + 2 \,{\left (b^{3} c - a b^{2} d\right )} x^{3}\right )} \sqrt{d x^{3} + c}}{3 \,{\left (a b^{4} c - a^{2} b^{3} d +{\left (b^{5} c - a b^{4} d\right )} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(d*x^3+c)^(1/2)/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

[-1/6*(((2*b^2*c - 3*a*b*d)*x^3 + 2*a*b*c - 3*a^2*d)*sqrt(b^2*c - a*b*d)*log((b*d*x^3 + 2*b*c - a*d + 2*sqrt(d
*x^3 + c)*sqrt(b^2*c - a*b*d))/(b*x^3 + a)) - 2*(3*a*b^2*c - 3*a^2*b*d + 2*(b^3*c - a*b^2*d)*x^3)*sqrt(d*x^3 +
c))/(a*b^4*c - a^2*b^3*d + (b^5*c - a*b^4*d)*x^3), 1/3*(((2*b^2*c - 3*a*b*d)*x^3 + 2*a*b*c - 3*a^2*d)*sqrt(-b
^2*c + a*b*d)*arctan(sqrt(d*x^3 + c)*sqrt(-b^2*c + a*b*d)/(b*d*x^3 + b*c)) + (3*a*b^2*c - 3*a^2*b*d + 2*(b^3*c
- a*b^2*d)*x^3)*sqrt(d*x^3 + c))/(a*b^4*c - a^2*b^3*d + (b^5*c - a*b^4*d)*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5} \sqrt{c + d x^{3}}}{\left (a + b x^{3}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(d*x**3+c)**(1/2)/(b*x**3+a)**2,x)

[Out]

Integral(x**5*sqrt(c + d*x**3)/(a + b*x**3)**2, x)

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Giac [A]  time = 1.10802, size = 150, normalized size = 1.1 \begin{align*} \frac{\frac{\sqrt{d x^{3} + c} a d^{2}}{{\left ({\left (d x^{3} + c\right )} b - b c + a d\right )} b^{2}} + \frac{2 \, \sqrt{d x^{3} + c} d}{b^{2}} + \frac{{\left (2 \, b c d - 3 \, a d^{2}\right )} \arctan \left (\frac{\sqrt{d x^{3} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{\sqrt{-b^{2} c + a b d} b^{2}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(d*x^3+c)^(1/2)/(b*x^3+a)^2,x, algorithm="giac")

[Out]

1/3*(sqrt(d*x^3 + c)*a*d^2/(((d*x^3 + c)*b - b*c + a*d)*b^2) + 2*sqrt(d*x^3 + c)*d/b^2 + (2*b*c*d - 3*a*d^2)*a
rctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^2))/d